I've a straight stick which has been broken into three randomlength pieces.
What is the probability that the pieces can be put together to form a triangle?
If you can answer this at this point, please do.
If not, perhaps this will help: here are several methods to break the stick into the three random length pieces:
 I select, independently, and at random, two points from the points that range uniformly along the stick, then break the stick at these two points.
 I select one point, independently, and at random (again uniformly), and break the stick at this point. I then randomly (with even chances) select one of the two sticks and randomly select a point (again uniformly) along that stick, and break it at that point.
 I select one point, independently, and at random (again uniformly), and break the stick at this point. I then select the larger stick, and randomly select a point (again uniformly) along that stick, and break it at that point.
If this clarifies the problem, please show how this affects your work.
(In reply to
I think this is a solution by puzzlesrfun)
"The sides of these smaller triangles are given by sqrt(2)*(1/4)*L. The 1/4 follows from the fact that two of the components are less than or equal to (1/4)*L, and therefore must sum to less than (1/2)*L."
But, actually, two of the components can add up to less than (1/2)*L even if not both of them are less than (1/4)*L. For instance, one is (1/8)*L and the other is (5/16)*L: the total is still less than half L. The better way to look at it is that no one component can be more than (1/2) L. The side of each small triangle is sqrt(2)*L/2.
As the larger triangle has area (sqrt(3)/2)*L^2 and the small triangles have area (sqrt(3)/2)*(L/2)^2, we get
(sqrt(3)/2)*L^2  3*(sqrt(3)/2)*(L/2)^2
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(sqrt(3)/2)*L^2
or 1  3/4 = 1/4.

Posted by Charlie
on 20031210 14:03:03 